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katen-ka-za [31]
3 years ago
13

Please help i will give brainliest

Mathematics
1 answer:
arlik [135]3 years ago
6 0
Here’s my answers, for #10 it got cut off but I got m=9/7 and d=11.40

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Which equation does the graph of the systems of equations solve?
DiKsa [7]

Answer:

-1/3x+3 = x-1

Step-by-step explanation:

The solution is (3,-2)

Check and see if the point solves the equation

-1/3x+3 = x-1

-1/3(3) +3 = 3-1

-1+3 = 3-1

2=2 yes

8 0
3 years ago
25 POINTS FOR ANSWER<br> 7x+5y=19<br> -7x-2y=-16
pashok25 [27]
Your answer for this problem would be 76-89+ 38, which equals negative 48
3 0
3 years ago
243^-y=(1/243)^3y*9^-2y<br><br><br> a.) y=-1<br> b.) y=0<br> c.) y=1<br> d.) no solution
allochka39001 [22]
243^{-y}=(\frac{1}{243})^{3y}\times 9^{-2y} has y = 0 as a solution.

Selection B is appropriate.
8 0
3 years ago
Read 2 more answers
Does anyone know how to do this?? Help please!!!!
Doss [256]

Answer:

When we have a rational function like:

r(x) = \frac{x + 1}{x^2 + 3}

The domain will be the set of all real numbers, such that the denominator is different than zero.

So the first step is to find the values of x such that the denominator (x^2 + 3) is equal to zero.

Then we need to solve:

x^2 + 3 = 0

x^2 = -3

x = √(-3)

This is the square root of a negative number, then this is a complex number.

This means that there is no real number such that x^2 + 3 is equal to zero, then if x can only be a real number, we will never have the denominator equal to zero, so the domain will be the set of all real numbers.

D: x ∈ R.

b) we want to find two different numbers x such that:

r(x) = 1/4

Then we need to solve:

\frac{1}{4} = \frac{x + 1}{x^2 + 3}

We can multiply both sides by (x^2 + 3)

\frac{1}{4}*(x^2 + 3) = \frac{x + 1}{x^2 + 3}*(x^2 + 3)

\frac{x^2 + 3}{4} = x + 1

Now we can multiply both sides by 4:

\frac{x^2 + 3}{4}*4 = (x + 1)*4

x^2 + 3 = 4*x + 4

Now we only need to solve the quadratic equation:

x^2 + 3 - 4*x - 4 = 0

x^2 - 4*x - 1 = 0

We can use the Bhaskara's formula to solve this, remember that for an equation like:

a*x^2 + b*x + c = 0

the solutions are:

x = \frac{-b +- \sqrt{b^2 - 4*a*c} }{2*a}

here we have:

a = 1

b = -4

c = -1

Then in this case the solutions are:

x = \frac{-(-4) +- \sqrt{(-4)^2 - 4*1*(-1)} }{2*(1)} = \frac{4 +- 4.47}{2}

x = (4 + 4.47)/2 = 4.235

x = (4 - 4.47)/2 = -0.235

5 0
3 years ago
Which triangle could be drawn as is it described
nignag [31]

Answer:

you didnt add the attachment

6 0
3 years ago
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